A thin rod of length 'L' lies along the x-axi | vedclass

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(A) The center of mass $x_{CM}$ of a rod with linear mass density $\lambda(x)$ is given by:$x_{CM} = \frac{\int_0^L x \lambda(x) dx}{\int_0^L \lambda(

Vedclass · Accepted Answer

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(A) The center of mass $x_{CM}$ of a rod with linear mass density $\lambda(x)$ is given by:$x_{CM} = \frac{\int_0^L x \lambda(x) dx}{\int_0^L \lambda(x) dx}$Substituting $\lambda(x) = k(\frac{x}{L})^n$:$x_{CM} = \frac{\int_0^L x \cdot k(\frac{x}{L})^n dx}{\int_0^L k(\frac{x}{L})^n dx} = \frac{\frac{k}{L^n} \int_0^L x^{n+1} dx}{\frac{k}{L^n} \int_0^L x^n dx} = \frac{[\frac{x^{n+2}}{n+2}]_0^L}{[\frac{x^{n+1}}{n+1}]_0^L} = \frac{L^{n+2}/(n+2)}{L^{n+1}/(n+1)} = L \frac{n+1}{n+2}$Analyzing the function $f(n) = L \frac{n+1}{n+2} = L (1 - \frac{1}{n+2})$:$1$. For $n = 0$,$x_{CM} = L(1 - 1/2) = L/2$.$2$. As $n 	o \infty$,$x_{CM} 	o L$.$3$. The derivative $\frac{dx_{CM}}{dn} = L \frac{(n+2) - (n+1)}{(n+2)^2} = \frac{L}{(n+2)^2} > 0$,so $x_{CM}$ is an increasing function.$4$. The second derivative $\frac{d^2x_{CM}}{dn^2} = -\frac{2L}{(n+2)^3} These properties match the graph in option $(A)$.

Quantities	Their numerical values (in $SI$ units)
a. Angular acceleration of smaller wheel	$1$. $5/3$
b. Torque on the larger wheel	$2$. $100/3$
c. Angular acceleration of larger wheel	$3$. $5/2$

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